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:<math>f_a: G \to G, \quad f_a(x) = x^{-1}a\sigma(x)</math>.
Then we have:
:<math>(d f_a)_e = d(h \circ (x \mapsto (x^{-1}, a, \sigma(x))))_e = dh_{(e,
where <math>h(x, y, z) = xyz</math>. It follows <math>(d f_a)_e</math> is bijective since the differential of the Frobenius <math>\sigma</math> vanishes. Since <math>f_a(bx) = f_{f_a(b)}(x)</math>, we also see that <math>(df_a)_b</math> is bijective for any ''b''. Let ''X'' be the closure of the image of <math>f_1</math>. Then the smooth points of ''X'' form an open dense subset of ''X''; thus, there is some ''b'' in ''G'' such that <math>f_1(b)</math> is a smooth point of ''X''. Since the tangent space to ''X'' at <math>f_1(b)</math> and the tangent space to ''G'' at ''b'' have the same dimension, it follows that ''X'' and ''G'' have the same dimension, since ''G'' is smooth. Since ''G'' is connected, the image of <math>f_1</math> then contains an open dense subset ''U'' of ''G''. Now, given an element ''a'' in ''G'', for the same reasoning, the image of <math>f_a</math> contains an open dense subset ''V'' of ''G''. The intersection <math>U \cap V</math> is then nonempty but then this implies ''a'' is in the image of <math>f_1</math>.
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